Question
Find the general solution of $\sin 2 x+\sin 4 x+\sin 6 x=0$
$(\sin 2 x+\sin 6 x)+\sin 4 x=0$
$2 \sin 4 x \cdot \cos 2 x+\sin 4 x=0$
$\sin 4 x(2 \cos 2 x+1)=0$
$\sin 4 x=0$ or $2 \cos 2 x+1=0$
$\sin 4 x=0$ or $\cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)$
Using $\sin x=0 \Rightarrow x=n \pi$
$\sin 4 x=0$
$4 x=n \pi$
The genral solution is $x$
$x=\frac{n \pi}{4}$
using $\cos x=\cos \alpha \Rightarrow x=2 m x \pm \alpha$
$\cos 2 x=\cos \left(\frac{2 \pi}{3}\right)$
$2 x=2 \mp i \pm \frac{2 \pi}{3}$
The genral solution is x
$x=\mp i \pm \frac{\pi}{3}$ where $m, n \in z$
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